Mixing
Example of measure for some algebra
Can you give an example of a finitely additive measure given on $mathcal A$, not
having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?
probability-theory measure-theory
asked Nov 20 '18 at 10:12
Lisa
255
add a comment |
Can you give an example of a finitely additive measure given on $mathcal A$, not
having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?
probability-theory measure-theory
asked Nov 20 '18 at 10:12
Lisa
255
add a comment |
Can you give an example of a finitely additive measure given on $mathcal A$, not
having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?
probability-theory measure-theory
asked Nov 20 '18 at 10:12
Lisa
255
Can you give an example of a finitely additive measure given on $mathcal A$, not
having a countably additive extension to a $sigma$-algebra generated by $mathcal A$ for some algebra $mathcal A$ over $mathbb N$?
probability-theory measure-theory
probability-theory measure-theory
asked Nov 20 '18 at 10:12
Lisa
255
asked Nov 20 '18 at 10:12
Lisa
255
asked Nov 20 '18 at 10:12
Lisa
255
asked Nov 20 '18 at 10:12
Lisa
255
asked Nov 20 '18 at 10:12
Lisa
255
255
add a comment |
add a comment |
1 Answer
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On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
add a comment |
On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
add a comment |
On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
On $mathbb N$ with the sigma algebra of all subsets let $mu (A)=0$ if $A$ is a finite set and $infty$ if it is an infinite set. Then $mu (mathbb N)=infty$ and $sum _n mu({n})=0$ so $mu$ is not countably additive. It is trivial to check that $mu$ is finitely additive.
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 10:16
Kavi Rama Murthy
50.4k31854
50.4k31854
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
add a comment |
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
Thank you, it helps me
– Lisa
Nov 20 '18 at 10:34
add a comment |
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